Question ID: #490
Let $[x]$ denote the greatest integer less than or equal to $x$. Then domain of $f(x)=\sec^{-1}(2[x]+1)$ is:
- (1) $(-\infty,-1]\cup[0,\infty)$
- (2) $(-\infty,\infty)$
- (3) $(-\infty,-1]\cup[1,\infty)$
- (4) $(-\infty,\infty)-\{0\}$
Solution:
The domain of $\sec^{-1}(t)$ is $t \in (-\infty, -1] \cup [1, \infty)$.
Here $t = 2[x] + 1$.
So, we need $2[x] + 1 \le -1$ or $2[x] + 1 \ge 1$.
Case 1:
$$2[x] + 1 \le -1$$
$$2[x] \le -2$$
$$[x] \le -1$$
This implies $x < 0$. So $x \in (-\infty, 0)$.
Case 2:
$$2[x] + 1 \ge 1$$
$$2[x] \ge 0$$
$$[x] \ge 0$$
This implies $x \ge 0$. So $x \in [0, \infty)$.
Combining both cases:
Domain $= (-\infty, 0) \cup [0, \infty) = (-\infty, \infty)$.
Ans. (2)
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